# Charge operator applied to matrix multiplets

In the context of SM ($$SU(3)_C\otimes SU(2)_L\otimes U(1)_Y$$) the charge operator is $$Q_{SM} = T_3 + \frac{Y}{2}\mathbb{I}_2$$ and gives us the fermions charges. Here $$T_3=\frac{1}{2}\sigma_3$$ is the third $$SU(2)$$ generator.

For example, assuming $$Y=-1$$ for the left lepton doublet, $$Q_{SM}\Psi=\begin{pmatrix}0&0\\0&-1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L\end{pmatrix}$$ and the charge components are obtained.

This process works fine for the scalar doublet too, $$Q_{SM}\begin{pmatrix}\phi^+\\\phi^0\end{pmatrix}=\begin{pmatrix}+1 \phi^+\\0 \phi^0\end{pmatrix}$$, with $$Y=+1$$.

On the other hand, the same calculation can be applied for $$SU(3)_C\otimes SU(3)_L\otimes U(1)_X$$ extension in which the charge operator is $$Q_{331} = T_3-\sqrt{3}T_8+X\mathbb{I}_3$$ where $$T_3=\frac{1}{2}\lambda_3$$ and $$T_8=\frac{1}{2}\lambda_8$$ are the $$SU(3)$$ diagonal generators.

For example, $$Q_{331}\Psi_1=\begin{pmatrix}0&0&0\\0&-1&0\\0&0&1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L \\e_L^c\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L \\ +1 e_L^c\end{pmatrix}$$ and the charge components are obtained again with $$X=0$$.

In the same way as before, we can do this with the scalar multiplets. For example with one of the triplets called $$\eta$$, $$Q_{331}\eta=\begin{pmatrix}1&0&0\\0&0&0\\0&0&2\end{pmatrix}\begin{pmatrix}\eta^+ \\ \eta^0 \\ \eta^{++}\end{pmatrix}=\begin{pmatrix}+1 \eta^+ \\ 0 \eta^0 \\ +2 \eta^{++}\end{pmatrix}$$ and the charge components are obtained again with $$X=+1$$.

The question is, ¿what about the matrix scalar multiplets? e.g. for a 331 sextet defined as $$S=\begin{pmatrix} \sigma_1^0&h_2^-&h_1^+ \\ h_2^- & H_1^{--} & \sigma_2^0 \\ h_1^+ & \sigma_2^0 & H_2^{++}\end{pmatrix},$$ with hypercharge $$X=0$$.

When you try to do the same process, but with $$Q_{331}^{\dagger}SQ_{331}$$ for being $$S$$ a matrix and not a vector, i can't obtain the charges. Here $$Q_{331}$$ is the same as for $$\Psi_1$$.

¿Any idea in order to obtain the component charges?

• What is the representation of the sextet $S$ under the gauge group? – InertialObserver Feb 21 at 18:04
• In the group $SU(3)_C\otimes SU(3)_L\otimes U(1)_X$ which contains SM group, the sextet transforms as $S\sim (1_C,6_L,0)$. That is $S$ is a singlet in $SU(3)_C$, a sextet in $SU(3)_L$ with hypercharge zero. – Eduardo Castillo Ruiz Feb 21 at 23:21